Mysql经典50道SQL练习题及答案与详细分析_数据库原理50题答案解析

数据准备

• 1.学生表
  Student(SId,Sname,Sage,Ssex)
  –SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-01-01' , '女');
insert into Student values('07' , '郑竹' , '1989-01-01' , '女');
insert into Student values('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2012-06-06' , '女');
insert into Student values('12' , '赵六' , '2013-06-13' , '女');
insert into Student values('13' , '孙七' , '2014-06-01' , '女');
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• 2.课程表
  Course(CId,Cname,TId)
  –CId 课程编号,Cname 课程名称,TId 教师编号

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
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• 3.教师表
  Teacher(TId,Tname)
  –TId 教师编号,Tname 教师姓名

create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
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• 4.成绩表
  SC(SId,CId,score)
  –SId 学生编号,CId 课程编号,score 分数

create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
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练习题目

查询" 01 “课程比” 02 "课程成绩高的学生的信息及课程分数

因为需要全部的学生信息，则需要在sc表中得到符合条件的SId后与student表进行join，可以left join 也可以 right join

select * from Student RIGHT JOIN (
    select t1.SId, class1, class2 from
          (select SId, score as class1 from sc where sc.CId = '01')as t1, 
          (select SId, score as class2 from sc where sc.CId = '02')as t2
    where t1.SId = t2.SId AND t1.class1 > t2.class2
)r 
on Student.SId = r.SId;
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select * from  (
    select t1.SId, class1, class2 
    from
        (SELECT SId, score as class1 FROM sc WHERE sc.CId = '01') AS t1, 
        (SELECT SId, score as class2 FROM sc WHERE sc.CId = '02') AS t2
    where t1.SId = t2.SId and t1.class1 > t2.class2
) r 
LEFT JOIN Student
ON Student.SId = r.SId;
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查询同时存在" 01 “课程和” 02 "课程的成绩情况

select * from 
    (select * from sc where sc.CId = '01') as t1, 
    (select * from sc where sc.CId = '02') as t2
where t1.SId = t2.SId;
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查询存在" 01 “课程但可能不存在” 02 "课程的成绩情况(不存在时显示为 null )

这一道就是明显需要使用join的情况了，02可能不存在，即为left join的右侧或right join 的左侧即可.

select * from 
(select * from sc where sc.CId = '01') as t1
left join 
(select * from sc where sc.CId = '02') as t2
on t1.SId = t2.SId;
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select * from 
(select * from sc where sc.CId = '02') as t2
right join 
(select * from sc where sc.CId = '01') as t1
on t1.SId = t2.SId;
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查询不存在" 01 “课程但存在” 02 "课程的成绩情况

select * from sc
where sc.SId not in (
    select SId from sc 
    where sc.CId = '01'
) 
AND sc.CId= '02';
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查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

这里只用根据学生ID把成绩分组，对分组中的score求平均值，最后在选取结果中AVG大于60的即可. 注意，这里必须要给计算得到的AVG结果一个alias.（AS ss）
得到学生信息的时候既可以用join也可以用一般的联合搜索

select student.SId,sname,ss from student,(
    select SId, AVG(score) as ss from sc  
    GROUP BY SId 
    HAVING AVG(score)> 60
    )r
where student.sid = r.sid;
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select Student.SId, Student.Sname, r.ss from Student right join(
      select SId, AVG(score) AS ss from sc
      GROUP BY SId
      HAVING AVG(score)> 60
)r on Student.SId = r.SId;
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select s.SId,ss,Sname from(
select SId, AVG(score) as ss from sc  
GROUP BY SId 
HAVING AVG(score)> 60
)r left join 
(select Student.SId, Student.Sname from
Student)s on s.SId = r.SId;
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查询在 SC 表存在成绩的学生信息

select DISTINCT student.*
from student,sc
where student.SId=sc.SId
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查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

联合查询不会显示没选课的学生：

select student.sid, student.sname,r.coursenumber,r.scoresum
from student,
(select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc 
group by sc.sid)r
where student.sid = r.sid;
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如要显示没选课的学生(显示为NULL)，需要使用left join:

select s.sid, s.sname,r.coursenumber,r.scoresum
from (
    (select student.sid,student.sname 
    from student
    )s 
    left join 
    (select 
        sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber
        from sc 
        group by sc.sid
    )r 
   on s.sid = r.sid
);
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查有成绩的学生信息

这一题涉及到in和exists的用法，在这种小表中，两种方法的效率都差不多，in和exists的具体区别分析请参考https://www.jianshu.com/p/f212527d76ff，
当表2的记录数量非常大的时候，选用exists比in要高效很多.EXISTS用于检查子查询是否至少会返回一行数据，该子查询实际上并不返回任何数据，而是返回值True或False.
结论：IN()适合B表比A表数据小的情况
结论：EXISTS()适合B表比A表数据大的情况

select * from student 
where exists (select sc.sid from sc where student.sid = sc.sid);
#或者
select * from student 
where exists (select 1 from sc where student.sid = sc.sid);
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select * from student
where student.sid in (select sc.sid from sc);
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查询「李」姓老师的数量

select count(*)
from teacher
where tname like '李%';
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查询学过「张三」老师授课的同学的信息

多表联合查询

select student.* from student,teacher,course,sc
where 
    student.sid = sc.sid 
    and course.cid=sc.cid 
    and course.tid = teacher.tid 
    and tname = '张三';
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查询没有学全所有课程的同学的信息

因为有学生什么课都没有选，反向思考，先查询选了所有课的学生，再选择这些人之外的学生.

select * from student
where student.sid not in (
  select sc.sid from sc
  group by sc.sid
  having count(sc.cid)= (select count(cid) from course)
);
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查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息

这个用联合查询也可以，但是逻辑不清楚，我觉得较为清楚的逻辑是这样的：从sc表查询01同学的所有选课cid–从sc表查询所有同学的sid如果其cid在前面的结果中–从student表查询所有学生信息如果sid在前面的结果中

select * from student 
where student.sid in (
    select sc.sid from sc 
    where sc.cid in(
        select sc.cid from sc 
        where sc.sid = '01'
    )
);
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查询和" 01 "号的同学学习的课程 完全相同的其他同学的信息

SELECT
 Student.*
FROM
 Student
WHERE
 s_id IN (select s_id from score GROUP BY s_id HAVING GROUP_CONCAT(c_id) = (    
     SELECT GROUP_CONCAT(c_id) FROM Score WHERE s_id = '01')
 );
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查询没学过"张三"老师讲授的任一门课程的学生姓名

仍然还是嵌套，三层嵌套， 或者多表联合查询

select * from student
    where student.sid not in(
        select sc.sid from sc where sc.cid in(
            select course.cid from course where course.tid in(
                select teacher.tid from teacher where tname = "张三"
            )
        )
    );
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select * from student
where student.sid not in(
    select sc.sid from sc,course,teacher 
    where
        sc.cid = course.cid
        and course.tid = teacher.tid
        and teacher.tname= "张三"
);
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查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

select student.SId, student.Sname,b.avg
from student RIGHT JOIN
(select sid, AVG(score) as avg from sc
    where sid in (
              select sid from sc 
              where score<60 
              GROUP BY sid 
              HAVING count(score)>1)
    GROUP BY sid) b on student.sid=b.sid;
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检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select student.*, sc.score from student, sc
where student.sid = sc.sid
and sc.score < 60
and cid = "01"
ORDER BY sc.score DESC;
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按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select *  from sc 
left join (
    select sid,avg(score) as avscore from sc 
    group by sid
    )r 
on sc.sid = r.sid
order by avscore desc;
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查询各科成绩最高分、最低分和平均分：

以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率

及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列

select 
sc.CId ,
max(sc.score)as 最高分,
min(sc.score)as 最低分,
AVG(sc.score)as 平均分,
count(*)as 选修人数,
sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,
sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率 
from sc
GROUP BY sc.CId
ORDER BY count(*)DESC, sc.CId ASC
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按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

这一道题有点tricky，可以用变量，但也有更为简单的方法，即自交（左交）
用sc中的score和自己进行对比，来计算“比当前分数高的分数有几个”。
按各科成绩进行排序，并显示排名， Score 重复时合并名次.

select a.cid, a.sid, a.score, count(b.score)+1 as rank
from sc as a 
left join sc as b 
on a.score<b.score and a.cid = b.cid
group by a.cid, a.sid,a.score
order by a.cid, rank ASC;
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查询学生的总成绩，并进行排名，总分重复时保留名次空缺

这里主要学习一下使用变量。在SQL里面变量用@来标识。

set @crank=0;
select q.sid, total, @crank := @crank +1 as rank from(
select sc.sid, sum(sc.score) as total from sc
group by sc.sid
order by total desc)q;
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查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比

有时候觉得自己真是死脑筋。group by以后的查询结果无法使用别名，所以不要想着先单表group by计算出结果再从第二张表里添上课程信息，而应该先将两张表join在一起得到所有想要的属性再对这张总表进行统计计算。这里就不算百分比了，道理相同。
注意一下，用case when 返回1 以后的统计不是用count而是sum

select course.cname, course.cid,
sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end) as "[100-85]",
sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end) as "[85-70]",
sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end) as "[70-60]",
sum(case when sc.score<=60 and sc.score>0 then 1 else 0 end) as "[60-0]"
from sc left join course
on sc.cid = course.cid
group by sc.cid;
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查询各科成绩前三名的记录

计算比自己分数大的记录有几条，如果小于3 就select，因为对前三名来说不会有3个及以上的分数比自己大了，最后再对所有select到的结果按照分数和课程编号排名即可。

select * from sc
where (
select count(*) from sc as a 
where sc.cid = a.cid and sc.score<a.score 
)< 3
order by cid asc, sc.score desc;
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查询每门课程被选修的学生数

select cid, count(sid) from sc 
group by cid;
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查询出只选修两门课程的学生学号和姓名

select student.sid, student.sname from student
where student.sid in
(select sc.sid from sc
group by sc.sid
having count(sc.cid)=2
);
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或 联合查询

select student.SId,student.Sname
from sc,student
where student.SId=sc.SId  
GROUP BY sc.SId
HAVING count(*)=2；
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查询男生、女生人数

select ssex, count(*) from student
group by ssex
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查询名字中含有「风」字的学生信息

select *
from student 
where student.Sname like '%风%'
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查询同名同性学生名单，并统计同名人数

select sname, count(*) from student
group by sname
having count(*)>1;
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嵌套查询列出同名的全部学生的信息

select * from student
where sname in (
select sname from student
group by sname
having count(*)>1
);
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查询 1990 年出生的学生名单

select *
from student
where YEAR(student.Sage)=1990;
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查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

select sc.cid, course.cname, AVG(SC.SCORE) as average from sc, course
where sc.cid = course.cid
group by sc.cid 
order by average desc,cid asc;
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查询平均成绩大于等于 85 的所有学生的学号、姓名和平均成绩

select student.sid, student.sname, AVG(sc.score) as aver from student, sc
where student.sid = sc.sid
group by sc.sid
having aver > 85;
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查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

select student.sname, sc.score from student, sc, course
where student.sid = sc.sid
and course.cid = sc.cid
and course.cname = "数学"
and sc.score < 60;
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查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

select student.sname, cid, score from student
left join sc
on student.sid = sc.sid;
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查询任何一门课程成绩在 70 分以上的姓名、课程名称和分数

select student.sname, course.cname,sc.score from student,course,sc
where sc.score>70
and student.sid = sc.sid
and sc.cid = course.cid;
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查询不及格的课程

取唯一可以用group by ，也可以用distinct

select cid from sc
where score< 60
group by cid;
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select DISTINCT sc.CId
from sc
where sc.score <60;
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查询课程编号为 01 且课程成绩在 80 分以上的学生的学号和姓名

select student.sid,student.sname 
from student,sc
where cid="01"
and score>=80
and student.sid = sc.sid;
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求每门课程的学生人数

select sc.CId,count(*) as 学生人数
from sc
GROUP BY sc.CId;
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成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

用having max()理论上也是对的，但是下面那种按分数排序然后取limit 1的更直观可靠

select student.*, sc.score, sc.cid from student, teacher, course,sc 
where teacher.tid = course.tid
and sc.sid = student.sid
and sc.cid = course.cid
and teacher.tname = "张三"
having max(sc.score);
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取limit 1的更直观可靠

select student.*, sc.score, sc.cid from student, teacher, course,sc 
where teacher.tid = course.tid
and sc.sid = student.sid
and sc.cid = course.cid
and teacher.tname = "张三"
order by score desc
limit 1;
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成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

为了验证这一题，先修改原始数据

UPDATE sc SET score=90
where sid = "07"
and cid ="02";
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这样张三老师教的02号课就有两个学生同时获得90的最高分了。
这道题的思路继续上一题，我们已经查询到了符合限定条件的最高分了，这个时候只用比较这张表，找到全部score等于这个最高分的记录就可，看起来有点繁复。

select student.*, sc.score, sc.cid from student, teacher, course,sc 
where teacher.tid = course.tid
and sc.sid = student.sid
and sc.cid = course.cid
and teacher.tname = "张三"
and sc.score = (
    select Max(sc.score) 
    from sc,student, teacher, course
    where teacher.tid = course.tid
    and sc.sid = student.sid
    and sc.cid = course.cid
    and teacher.tname = "张三"
);
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查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩

同上，在这里用了inner join后会有概念是重复的记录：“01 课与 03课”=“03 课与 01 课”，所以这里取唯一可以直接用group by

select  a.cid, a.sid,  a.score from sc as a
inner join 
sc as b
on a.sid = b.sid
and a.cid != b.cid
and a.score = b.score
group by cid, sid;
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查询每门功成绩最好的前两名

select a.sid,a.cid,a.score from sc as a 
left join sc as b 
on a.cid = b.cid and a.score<b.score
group by a.cid, a.sid
having count(b.cid)<2
order by a.cid;
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统计每门课程的学生选修人数（超过 5 人的课程才统计）。

select sc.cid, count(sid) as cc from sc
group by cid
having cc >5;
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检索至少选修两门课程的学生学号

select sid, count(cid) as cc from sc
group by sid
having cc>=2;
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查询选修了全部课程的学生信息

select student.*
from sc ,student 
where sc.SId=student.SId
GROUP BY sc.SId
HAVING count(*) = (select DISTINCT count(*) from course )
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查询各学生的年龄，只按年份来算

按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

select student.SId as 学生编号,student.Sname  as  学生姓名,
TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄
from student
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查询本周过生日的学生

select *
from student 
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());
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查询下周过生日的学生

select *
from student 
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;
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查询本月过生日的学生

select *
from student 
where MONTH(student.Sage)=MONTH(CURDATE());
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查询下月过生日的学生

select *
from student 
where MONTH(student.Sage)=MONTH(CURDATE())+1;
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原文：https://www.jianshu.com/p/476b52ee4f1b
https://blog.csdn.net/fashion2014/article/details/78826299